f(x)=[x^2+5]/[x-3] where x not equal to 3
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Originally Posted by write2rvr f(x)=[x^2+5]/[x-3] where x not equal to 3 set and make x the subject: <-- from completing the square Now set any y to x and set x to
Originally Posted by write2rvr f(x)=[x^2+5]/[x-3] where x not equal to 3 When you get to: . . . . .0 = x^2 - xy + 3y + 5 ...you can always plug this into the Quadratic Formula, using a = 1, b = -y, and c = 3y + 5.
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